Stat 219  Stochastic Processes
Homework Set 8, Fall 2009, Due: November 18th
1. Exercise 5.2.5.
Show that
E
(
τ
β,α
) =
αβ
by applying Doob’s optional stopping theorem for the uniformly
integrable stopped martingale
W
2
t
∧
τ
β,α

t
∧
τ
β,α
.
ANS:
We have seen enroute to (5.2.2) that
τ
β,α
≤
τ
α
<
∞
almost surely. Considering the martingale
X
t
=
W
2
t

t
of continuous sample path we have further assumed in the statement of the exercise that
X
t
∧
τ
β,α
is U.I. Thus, Doob’s optional stopping theorem (Theorem 4.3.17) applies here, leading to the
identity
E
(
W
2
τ
β,α

τ
β,α
) =
E
(
W
2
0

0) = 0. That is,
E
τ
β,α
=
E
W
2
τ
β,α
=
α
2
P
(
W
τ
β,α
=
α
) +
β
2
P
(
W
τ
β,α
=

β
) =
α
2
β
α
+
β
+
β
2
α
α
+
β
=
αβ.
2. Exercise 5.2.7
(a) Suppose a random vector
Y
is independent of random vectors
Z
1
, Z
2
and
Z
1
has the same law as
Z
2
. Show that then
Y
+
Z
1
has the same law as
Y
+
Z
2
.
ANS:
A simple computation using characteristic functions and independence yields,
E
exp(
i
(
θ
, Y
+
Z
1
)) =
E
exp(
i
(
θ
, Y
) +
i
(
θ
, Z
1
)) =
E
exp(
i
(
θ
, Y
))
E
exp(
i
(
θ
, Z
1
)) (independence)
=
E
exp(
i
(
θ
, Y
))
E
exp(
i
(
θ
, Z
2
)) (
Z
1
L
=
Z
2
)
=
E
exp(
i
(
θ
, Y
) +
i
(
θ
, Z
2
)) (independence)
=
E
exp(
i
(
θ
, Y
+
Z
2
))
.
We conclude that the characteristic functions of
Y
+
Z
1
and
Y
+
Z
2
agree and hence the two
random vectors have the same distribution.
(b) Suppose
{
Y
t
}
and
{
Z
t
}
are independent stochastic processes and
{
Z
t
}
has the same distribution
as
{
Z
t
}
. Explain why
X
t
=
Y
t

Z
t
has the same distribution as
W
t
=
Y
t
+
Z
t
.
ANS:
This immediately follows by applying part (a) to the finite dimensional distributions of
{
Y
t
}
,
{
Z
t
}
and
{
Z
t
}
.
(c) Let
W
t
be a Brownian motion and
τ
≥
0 nonrandom. Upon verifying that part (b) applies for
Y
t
=
W
t
∧
τ
and
Z
t
=
W
t

W
t
∧
τ
deduce that
X
τ
t
=
W
t
∧
τ

(
W
t

W
t
∧
τ
) is a Brownian motion.
ANS:
To apply (b) we need to show that
{
Y
t
}
and
{
Z
t
}
are independent stochastic processes. Let
{G
t
}
denote the canonical filtration of
{
W
t
}
and note that
τ
is (trivially) a stopping time for
{G
t
}
.
Note that
Y
t
is measurable with respect to
G
τ
for all
t
≥
0. Furthermore,
Z
t
= 0 when
t
≤
τ
and
Z
t
=
W
t

W
τ
for
t
≥
τ
. Hence it follows from Proposition 5.2.3 that
Z
t
is independent of
{G
τ
}
1
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and hence of
Y
t
. Therefore by part (b),
{
X
τ
t
}
=
{
Y
t

Z
t
}
L
=
{
Y
t
+
Z
t
}
=
{
W
t
}
,
so that
{
X
τ
t
}
has the same finite dimensional distributions as a Brownian motion. Since its sample
path
t
→
X
τ
t
is continuous, the desired result follows.
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 Spring '09
 Brownian Motion, Stochastic process, WT

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